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STRUCTURE OF BARIUM ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy). ( September 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories which cannot lead to the nuclear structure. Under this physics crisis in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838,68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications Naturally occurring barium (Ba) is a mix of six stable isotopes and one very long-lived radioactive primordial isotope, barium-130, recently identified as being unstable by geochemical means (from analysis of the presence of its daughter xenon-130 in rocks). This nuclide decays by doubling -electron capture (absorbing two electrons and emitting two neutrinos) with a half-life of (0.5-2.7)×1021 years (about 1011 times the age of the universe). There are a total of thirty-three known radioisotopes in addition to 130Ba, but most of these are highly radioactive with half-lives in the several millisecond to several minute range. The only notable exceptions are 133Ba which has a half-life of 10.51 years, 131Ba (11.5 days), and 137mBa (2.55 minutes), which is the decay product of 137Cs (30.17 years, and a common fission product). STRUCTURE OF Ba-114, Ba-116, Ba-118, Ba-120, Ba-122, Ba-124, Ba-126, Ba-128, Ba-130, Ba-132, Ba-134, Ba-136, AND Ba-138 WITH S = 0 For understanding the structure of the above nuclides having even number of extra neutrons you must read my STRUCTUTRE OF Ba-132 . Here we have an even number of extra neutrons which fill the blank positions of the structure of Ba-112 with S = 0. For example the unstable Ba-130 has 18 extra neutrons with opposite spins, which make two bonds per neutron but the small number of them cannot give enough binding energies to pn bonds for overcoming the pp and nn repulsions. However in the stable structures of Ba-132, Ba-134, Ba-136, and Ba-138 the greater number of extra neutrons with opposite spins gives enough binding energies to pn bonds for overcoming the repulsions. STRUCTURE OF Ba-140, Ba-142, Ba-144, Ba-146, Ba-148, Ba-150 AND Ba-152 WITH S =0 Similarly the structure of the above unstable nuclides with even number of extra neutrons is based on the structure of Ba-112 with S =0 . For example the Ba-140 has 28 extra neutrons of opposite spins but the two more extra neutrons than those of the stable Be-138, (in the absence of blank positions), make single bonds leading to the decay. Under this condition the more extra neutrons than those of Ba-138 in the above unstable nuclides make always single bonds leading to the decay. STRUCTURE OF Ba-117, Ba-125, Ba-127, Ba-129, Ba-131, Ba-133, Ba-135, Ba-137, AND Ba-147 For understanding the structure of the above nuclides with odd number of extra neutrons you must read my STRUCTURE OF Ba-135 . Here the structure of the above nuclides is based also on the same Ba-112 with S = 0, in which the odd number of extra neutrons gives S = +1/2 or S = +3/2. For example the unstable Ba-133 with S = + 1/2 of 21 extra neutrons has 11 extra neutrons of positive spins and 10 extra neutrons of negative spins giving S = +1/2. In other words the spin of Ba-133 with S = +1/2 is given by S = 0 + 11(+1/2) + 10(-1/2) = +1/2 . These extra neutrons make two bonds per neutron but the small number of them cannot give enough binding energies to pn bonds for overcoming the pp and nn repulsions. However in the stable Ba-135 and Ba-137 with S = +3/2 the greater number of extra neutrons gives enough binding energies to pn bonds for overcoming the repulsions. For example the Ba-137 with 25 extra neutrons has 14 extra neutrons of positive spins and 11 extra neutrons of negative spins able to give enough binding energies to pn bonds for overcoming the repulsions. However in the unstable Ba-147 with S = +3/2 the 10 more extra neutrons than those of the stable Ba-137, in the absence of blank positions, make single bonds leading to the decay. STRUCTURE OF Ba-115, Ba-119, Ba-121, AND Ba-123 In this group with odd number of extra neutrons we see that the structures are based on another structure of Ba-112 with S = +2. In the following diagram of Ba-112 with S = 0 we see that it turns to S = +2 when the p39n39 changes the spin from S =-1 to S = +1 giving S = +2. In this case it moves from the -HSQ to +HSQ for making bonds with p40n40. Then in the presence of the odd number of extra neutrons giving S = +1/2 one gets the structures of the above nuclides. For example the Ba-123 with S = +5/2 of 11 extra neutrons has 6 neutrons of positive spins and 5 extra neutrons of negative spins giving S = +1/2. That is S = +2 + 6(+1/2) + 5(-1/2) = +5/2 STRUCTURE OF Ba-139, Ba-141, Ba-143, Ba-145 Ba-149, Ba-151 AND Ba-153 This group with negative spins is based on another structure of Ba-112 having S =-2 . In the following diagram of Ba-112 with S = 0 we see that it has S = -2 when the p38n38 changes the spin from S = +1 to S = -1 giving S = -2. In this case it moves from the +HSQ to -HSQ for making bonds with p37n37. Then in the presence of the odd number of extra neutrons one gets the structures of the above nuclides. For example the Ba-153 with S = -5/2 of 41 extra neutrons has 20 extra neutrons of positive spins and 21 extra neutrons of negative spins. That is S = -2 +20(+1/2) + 21(-1/2) = -5/2 ' ' ' DIAGRAM OF Ba-112 WITH S = 0' In this structure of S = 0 based on Xe-108 with 54 protons and 54 neutrons you see the 6 horizontal planes of opposite spins like the +HP1, -HP2, +HP3, -HP4, +HP5 and -HP6 along the two horizontal squares, the -HSQ , and +HSQ. Here the p55n55 and p56n56 as vertical system s with S=0 are not shown. ' ' n40.......p40 ' +HSQ p38..........n38 ' ' n31………p12.........n12.......p32' ' -HP6 p31........n11.........p11…… n32 ' ' p29.........n10.........p10…… n30' ' +HP5 n29………p9..........n9 …….p30' ' p47.......n27.........p8............n8.........p28........n48' ' -HP4 n45.......p27.........n7..........p7.........n28..........p46 ' ' n47......p25.........n6.........p6..........n26...........p48' ' +HP3 p45......n25……….p5..........n5……….p26.........n46 ' ' n23……p4..........n4………….p24' ' -HP2 p23……....n3………p3………..n24 ' ' p21.........n2………p2...........n22' ' +HP1 n21........p1........n1.........p22' ' p37......n37 ' ' -HSQ n39.....p39' ' ' Category:Fundamental physics concepts